A photon and an electron, each of 20 eV energy, move in free space. The ratio of the linear momentum of the electron p_e to that of the photon p_ph, _ep_ph, is: [speed of light =3×10⁸ ms⁻¹, e=1.6×10⁻¹⁹ C, m_e=9×10⁻³¹ kg]
- A.(2)/(450)
- B.(1)/(250)
- C.225✓
- D.275
Correct Answer
(C) 225
Solution & Explanation
Compare momentum using the right relation for each. Electron (non-relativistic, KE = p²/2m): p_e = √(2m_e·E). Photon (massless, E = pc): p_ph = E/c. Take the ratio: p_e/p_ph = √(2m_e·E) ÷ (E/c) = c·√(2m_e/E). Convert energy: E = 20 eV = 20×1.6×10⁻¹⁹ = 3.2×10⁻¹⁸ J. Compute 2m_e/E = (2×9×10⁻³¹)/(3.2×10⁻¹⁸) = (18×10⁻³¹)/(3.2×10⁻¹⁸) = 5.625×10⁻¹³. √(5.625×10⁻¹³) = 7.5×10⁻⁷. p_e/p_ph = (3×10⁸)×(7.5×10⁻⁷) = 225. Trap: do NOT use p = E/c for the electron — that formula is only for the photon. The matched-energy electron carries far more momentum, giving p_e/p_ph = 225, option C.
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